Question

3.An empty 10ml volumetric flask weight 10.2634g when the flask is filled to the mark with and weighed again in the air at the mass 20.214g. what is the true volume of the flask at 25 degree celcius 4. How many liters of pure oxygen, measured at 740mmHg and 24 degree Celsius would be required to burn 1.00g of octane, C8H18 to carbon dioxide and water 5. when aluminum reacts with concentrated hydrochloric acid hydrogen evolves. The hydrogen is collected by displacement of water at 22 degrees celcius at a total pressure of 750 torr. It the volume of the collected gas is 365 ml, how many moles of hydrogen were collected? The vapor pressure of water at 22 degree celcius is 21 torr.

Answer 1

Let's tackle these questions one at a time.

### Problem 3: True Volume of the Flask
To find the true volume of the volumetric flask, we can first find the mass of the liquid that occupies the flask:

1. **Mass of the liquid:**
\[
\text{Mass of liquid} = \text{Mass of filled flask} - \text{Mass of empty flask}
\]
\[
\text{Mass of liquid} = 20.214 \, \text{g} - 10.2634 \, \text{g} = 9.9506 \, \text{g}
\]

2. **Density of the liquid:**
Assuming the liquid is water (or similar), the density at 25 °C is approximately \(1 \, \text{g/mL}\). Therefore, we can use the relation:
\[
\text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}}
\]
\[
\text{Volume} = \frac{9.9506 \, \text{g}}{1 \, \text{g/mL}} = 9.9506 \, \text{mL}
\]

### Problem 4: Liters of Pure Oxygen Required
The complete combustion of octane (\(C_8H_{18}\)) can be represented by the reaction:
\[
C_8H_{18} + 12.5 O_2 \rightarrow 8 CO_2 + 9 H_2O
\]

1. **Moles of Octane:**
\[
\text{Molar mass of } C_8H_{18} = (8 \times 12.01) + (18 \times 1.008) = 114.22 \, \text{g/mol}
\]

For 1.00 g of octane:
\[
\text{Moles of octane} = \frac{1.00 \, \text{g}}{114.22 \, \text{g/mol}} \approx 0.00875 \, \text{mol}
\]

2. **Moles of Oxygen Required:**
From the balanced equation, \( 1 \) mole of \(C_8H_{18}\) reacts with \( 12.5 \) moles of \(O_2\):
\[
\text{Moles of } O_2 = 12.5 \times 0.00875 \, \text{mol} \approx 0.109375 \, \text{mol}
\]

3. **Using Ideal Gas Law to Find Volume of O2:**
Using the Ideal Gas Law, \(PV = nRT\):
\[
n = 0.109375 \, \text{mol} \\
R = 0.0821 \, \text{L atm/(K mol)} \\
T = 24 °C + 273.15 = 297.15 \, K \\
P = 740 \, \text{mmHg} = \frac{740}{760} \, \text{atm} \approx 0.974 \, \text{atm}
\]

Rearranging the Ideal Gas Law:
\[
V = \frac{nRT}{P} = \frac{0.109375 \, \text{mol} \times 0.0821 \, \text{L atm/(K mol)} \times 297.15 \, K}{0.974 \, \text{atm}} \approx 2.69 \, \text{L}
\]

### Problem 5: Moles of Hydrogen Collected
1. **Correcting the Total Pressure:**
To find the pressure of the dry hydrogen gas, we need to subtract the vapor pressure of water from the total pressure:
\[
P_{H_2} = P_{\text{total}} - P_{\text{water}} = 750 \, \text{torr} - 21 \, \text{torr} = 729 \, \text{torr}
\]

2. **Convert pressure from torr to atm:**
\[
P_{H_2} = \frac{729 \, \text{torr}}{760} \approx 0.96 \, \text{atm}
\]

3. **Volume conversion:**
\[
V = 365 \, \text{mL} = 0.365 \, \text{L}
\]

4. **Using Ideal Gas Law to Find Moles of Hydrogen:**
\[
n = \frac{PV}{RT}
\]
\(R = 0.0821 \, \text{L atm/(K mol)}\) and \(T = 22 °C + 273.15 = 295.15 \, K\):
\[
n = \frac{0.96 \, \text{atm} \times 0.365 \, \text{L}}{0.0821 \, \text{L atm/(K mol)} \times 295.15 \, K} \approx 0.0142 \, \text{mol}
\]

### Summary of Answers:
- **True Volume of the Flask:** \(\approx 9.951 \, \text{mL}\)
- **Liters of Pure Oxygen Required:** \(\approx 2.69 \, \text{L}\)
- **Moles of Hydrogen Collected:** \(\approx 0.0142 \, \text{mol}\)